Question

A randomly selected sample of n =51 men in Brazil had an average lifespan of 59 years. The standard deviation was 10 years. Calculate a 98% confidence interval for the average lifespan for all men in Brazil.

Answer 1

`59-2.40(10)/(√(51))=55.639`

`59+ 2.40(10)/(√(51))=62.361`

The 98% confidence interval would be given by (55.639;62.361)

Explanation:

Information given

`\bar X= 59` represent the sample mean

`s= 10` represent the sample deviation

`n= 51` represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

The degrees of freedom are given by:

`df=n-1=51-1=50`

The Confidence is 0.98 or 98%, the significance would be
`\alpha=0.02` and
`\alpha/2 =0.01`,and the critical value would be
`t_(\alpha/2)=2.40`

And replacing we got:

`59-2.40(10)/(√(51))=55.639`

`59+ 2.40(10)/(√(51))=62.361`

The 98% confidence interval would be given by (55.639;62.361)